Help Understanding Energy, Distance, and Angled Shots

[jake_eizy]

Say my rangefinder, which accounts for angle, tells me the distance to the target is 410 yards. I estimate the angle I am shooting up the hill to be ~45 degrees, yes very steep. The true distance is then some where around 580 yards based on trig calculations. Now does my bullet impact the target with the Strelok energy estimates of 410 yards or 580 yards? Intuitively I would say 580 yards but I asked some of my aero space mechanical engineer friends who spent an hour or so writing out equations and got a value closer to 410 yards.

 

[wind gypsy]

Velocity (and thus energy) are going to be basically the same for a given LOS distance regardless of shot angle is my understanding.

 

[jake_eizy]

Velocity (and thus energy) are going to be basically the same for a given LOS distance regardless of shot angle is my understanding.

and LOS distance being the hypotenuse (580 yards) in me example above?

 

[Macintosh]

This is what my AB says for one of my rifles (note I did check your trig to verify the 580los=410 @ 45degrees, i.e. formula to find hypotenuse of a 90/45/45 triangle=410* sq root of 2=580ish)

580 yards @ 45degree look angle: 2.3mil elevation, impact velocity 1742 fps, energy 876 ft lb
410 yards @ 0degree look angle: 1.9mils elevation, impact velocity 2070 fps, energy 1237 ft lb
580 yards @ 0 degree look angle: 3.6mils elevation, impact velocity 1738 fps, energy 872 ft lb

The impact velocity figure lines up for me (which to me is much more important than energy) but I'm questioning my inputs given the difference in firing solution…the difference makes sense to me given more time for drag time/gravity to act on the bullet, but I had thought they should be closer at this yardage. I am wondering if someone else can double-check this, as I would have thought the elevation solution would have been +/- the same (or closer) between 410@ 0 degrees and 580 @ 45 degrees.

 

[BBob]

and LOS distance being the hypotenuse (580 yards) in me example above?

Yes. LOS velocity and TOF hardly change for the actual distance to the target so energy would be for 580.

 

[DiabeticKripple]

You shoot for the horizontal distance, not the total distance.

The 410yds would be correct

 

[BBob]

You shoot for the horizontal distance, not the total distance.

The 410yds would be correct

That is true but the bullet still travels 580 yards and still arrives at near the same velocity and TOF as it would if shot at 0° so to calculate energy it would be done with the velocity at target at 580.

 

[hereinaz]

Velocity degrades primarily as a result of air friction. That is LOS.

Wind deflection is from wind pushing the bullet over the whole flight. That is LOS.

Energy is a function of the velocity variation. That is LOS.

GRAVITY and only gravity acts over the angle compensated distance. Carefully picture what I describe below.

Gravity basically acts at 90 degrees to the center of the earth. Thus, a bullet shot at an angle is traveling a portion in the X axis and a portion in the Y axis. Gravity is calculated on the distance on the X axis that is perpendicular to gravity.

Gravity is so weak, that for a bullet going up at an angle and a bullet going down, there is very little difference on the Y axis.

It took me a lot to figure it out because most people don’t understand it and the internet is full of junk. It will only make sense once you understand how it all works.

 

[Macintosh]

@hereinaz does that mean the diff in firing solution I outlined above is an input error, or it is a small difference b/c gravity is acting on the bullet longer than if it were only TRAVELING 410 yards?

 

[hereinaz]

That is true but the bullet still travels 580 yards and still arrives at near the same velocity and TOF as it would if shot at 0° so to calculate energy it would be done with the velocity at target at 580.

This is the simple answer.

Your smart engineers did the word problem wrong…

 

[SirChooCH]

This should help

 

[hereinaz]

@hereinaz does that mean the diff in firing solution Inoutlined above is an input error, or it is a small difference b/c gravity is acting on the bullet longer than if it were only TRAVELING 410 yards?

In your 580 0 degree, gravity acts over the whole 580 because the bullet travels 580 yards over the X axis and zero up the Y axis. More drop.

In the 580 45degree, gravity acts over the only the portion it travels in the X axis. Every foot it travels over it travels some up. Needs less dope because gravity only acted on the X axis.

The difference between dope in the 580 45 degree and 410 0 degree happens because gravity also acts over time.

The bullet traveling 580 yards spends more time in the air so gravity has longer to act than it does on a bullet traveling 410 yards.

Gravity is a constant by time. So if you took the 580 yard angled shot and looked at the drop after the same time of flight as the 410 yard shot, it would have traveled 410 yards, it would be essentially the same drop.

 

[Macintosh]

Perfect, that matches my understanding—“longer”=for a longer amount of time. So my AB result is not likely to be from an input error.

 

[wind gypsy]

This is what my AB says for one of my rifles (note I did check your trig to verify the 580los=410 @ 45degrees, i.e. formula to find hypotenuse of a 90/45/45 triangle=410* sq root of 2=580ish)

580 yards @ 45degree look angle: 2.3mil elevation, impact velocity 1742 fps, energy 876 ft lb
410 yards @ 0degree look angle: 1.9mils elevation, impact velocity 2070 fps, energy 1237 ft lb
580 yards @ 0 degree look angle: 3.6mils elevation, impact velocity 1738 fps, energy 872 ft lb

The impact velocity figure lines up for me (which to me is much more important than energy) but I'm questioning my inputs given the difference in firing solution…the difference makes sense to me given more time for drag time/gravity to act on the bullet, but I had thought they should be closer at this yardage. I am wondering if someone else can double-check this, as I would have thought the elevation solution would have been +/- the same (or closer) between 410@ 0 degrees and 580 @ 45 degrees.

4 tenths seems like a lot between the first two scenarios but I've never spent much time taking shots at angles that steep!

 

[hereinaz]

If you want to read my explanation, it might help to associate it with all the bleeding on the drawing.

I probably should take this and develop it as an article in my weekly long range tips...

 

[hereinaz]

4 tenths seems like a lot between the first two scenarios but I've never spent much time taking shots at angles that steep!

A lot of this is hard to picture without the "sciency" framing, but it is correct.

But, we all know data is one thing, and shooting it is another. I invite everyone to try to shoot at a steep angle up or down a mountain. Almost guarantee that the dope will never match what you actually hit. Extreme angles introduce lots of influence from the change in position. The changes in centers of gravity significantly effect the recoil and influences on the rifle from the traditional flat shooting we do.

 

[jake_eizy]

A lot of this is hard to picture without the "sciency" framing, but it is correct.

But, we all know data is one thing, and shooting it is another. I invite everyone to try to shoot at a steep angle up or down a mountain. Almost guarantee that the dope will never match what you actually hit. Extreme angles introduce lots of influence from the change in position. The changes in centers of gravity significantly effect the recoil and influences on the rifle from the traditional flat shooting we do.

Thanks for your through explanation. Realistically the shot was most likely closer to 30 degrees but for ease of trig calc I used 45 degrees. I understand why the energy at impact is the LOS distance and not the distance from the range finder.

 

[MattB]

This is the simple answer.

Your smart engineers did the word problem wrong…

This

 

[hereinaz]

Thanks for your through explanation. Realistically the shot was most likely closer to 30 degrees but for ease of trig calc I used 45 degrees. I understand why the energy at impact is the LOS distance and not the distance from the range finder.

IMO, you should NEVER use the range corrected from the rangefinder and a ballistic calculator. It is too easy to be confused, even if you try to learn it. It took a lot for me to figure out how it worked. I had the benefit of really smart and educated shooters help me through it.

Ballistic calculators do all the math for you and rely on good inputs. Ballistic calculators are too easy and too common to rely on doing the math. I think everyone should learn the math and know how to use it, and have it available in their pack as a backup.

For most hunters, basically inside 400-450 yards hard dope and the angle compensated will work for drop. And, most wind calls will work with the compensated. At those ranges, it is close enough given the margin of errors are really inside the kill zone if you can make the shot without disturbing the rifle.

But, it also depends on the bullet and velocity, assuming a constant built in mathematical guesstimate. This is why the old school .270/25-06 are gonna have smaller groups than a 308/30-06. Initial velocity matters a lot in the first 500 yards. The BC of the bullet increasingly starts to matter more the further beyond 500.

 

[jake_eizy]

IMO, you should NEVER use the range corrected from the rangefinder and a ballistic calculator. It is too easy to be confused, even if you try to learn it. It took a lot for me to figure out how it worked. I had the benefit of really smart and educated shooters help me through it.

Ballistic calculators do all the math for you and rely on good inputs. Ballistic calculators are too easy and too common to rely on doing the math. I think everyone should learn the math and know how to use it, and have it available in their pack as a backup.

For most hunters, basically inside 400-450 yards hard dope and the angle compensated will work for drop. And, most wind calls will work with the compensated. At those ranges, it is close enough given the margin of errors are really inside the kill zone if you can make the shot without disturbing the rifle.

But, it also depends on the bullet and velocity, assuming a constant built in mathematical guesstimate. This is why the old school .270/25-06 are gonna have smaller groups than a 308/30-06. Initial velocity matters a lot in the first 500 yards. The BC of the bullet increasingly starts to matter more the further beyond 500.

So to give some context on my question, I was shooting a 6.5cm with Barnes 127 grain lrx bullet. I had created my dope chart out to 400 yards at 0 degrees based on StrelockPro app.

I ranged a cow elk at 410 yards, angle corrected. With estimated angle at 30 degrees, it was a steep canyon. Shot was a perfect lung shot with the bullet lodged in the opposite shoulder blade, she didn't go more then ten yards. I wanted to know energy as I was concerned my shot was ethically questionable on a cow elk. 410 range corrected at 30 degrees is LOS of ~475 yards. Which puts me at 1950 FPS and 1073 ft-lb.

So are you saying it is better to use LOS with my range finder then input range into Strelock with angle to get MOA adjustment? Along with possibly estimating angle in the field not using my phone. Seems may be challenging in the field. Either way 400-500 is near my max comfort level for shooting game.